Chemistry 125: Lecture 74
April 27, 2011
The Structure of Glucose
and
This
Synthesis of Two
Un-Natural Products
For copyright
notice see final
page of this file
“I may be allowed to denote
compounds [with] carbon plus
hydrogen and oxygen in the same
ratio that obtains in water…as
carbohydrates”
Schmidt (1844)
Grape Sugar (1792)
γλεῦκος
gleukos
“sweetness”
Glucose
Dumas (1838)
“ose” soonbecame the
Couper
generic suffix for sugars
1858
Carbohydrate
(C•HOH)n
CHOH
CHOH
HOH
CHOH
CHOH
CHOH
CHOH
aldose
ketose
(e.g. glucose)
(e.g. fructose)
Carbohydrate
(C•HOH)n
H
C=O
H
CHOH
H
CHOH
CHOH
CHOH
CHOH
C=O
CHOH
CHOH
C=O
CHOH
CHOH
CHOH
CHOH
CHOH
CHOH
CHOH
CHOH
CHOH
CHOH
CHOH
H
CHOH
H
CHOH
H
CHOH
We know the structure of glucose by spectroscopy
no aldehyde
or ketone!
http://riodb01.ibase.aist.go.jp/sdbs/
Crystalline Glucose
upon dissolution in D2O
[a]D = 112°
“dextrose”
1H-NMR
13C-NMR
After 1 day in D2O
no aldehyde
[a]D = 53°
no aldehyde
or ketone
J
7.9Hz
[a]D = 19°
anti H
H
hemiketal
with H+ or OH-
gauche
Cf. J. E. Gurst, J. Chem. Ed. 1003 (1991)
J
3.7Hz
We know the structure of glucose by spectroscopy
and X-ray Diffraction.
But how could they know?
Chirality - van’t Hoff (1877)
Fructose
Manna
Mannitol
H+ (1876)
reduced
HNO3
Mannose
(1888)
oxidized
Arabinose
(1873)
Gum
Arabic
Sucrose Galactose
Heinrich Kiliani
1855-1945
While a large number of compounds are very easily formed upon
oxidation of dextrose by dilute nitric acid or by halogens, these molecules
retain 6 carbon bound together in a chain. Under the same conditions
laevulose yields substances containing chains with a smaller number of
carbons (glycolic acid and inactive tartaric acid). Here oxidation causes
immediate splitting of the molecule, a fact which means that laevulose is a
ketone.1) Bearing in mind further the fact that laevulose in transformed
into mannitol by nascent hydrogen 2), one comes to the conclusion that
laevulose must be adjudged to have one of the following two constitutional
formulae:
Berlin
1885
One could hope to distinguish definitively between one and the other formula,
by succeeding in adding hydrocyanic acid to the ketone radical of laevulose and
transforming the cyanhydrin into the corresponding carboxylic acid. Since the
carboxylic acid from compound I above
?
“does not agree with the
KOH
HI / P
description
that Hecht
(STRONG)
gives of the Ca salt of
2) HCl
methylbutyacetic
acid,”
Ca, Ba,
Sr, Pb
(fuming)
salts all agree
must upon exhaustive reduction by concentrated hydriodic acid yield
methylbutylacetic acid, and on the contrary the carboxylic acid from
compound II under the same conditions leads to ethylpropylacetic acid.
?
1) HCN
1) HCN
NOT from Laevulose!HI / P
2) HCl
(fuming)
Kiliani
1886
“which means that
KOH
my heptanoic acid is
(STRONG)
identical with [unknown]
ethylpropylacetic acid.”
NOT ketone!
RCO2+ CH CO -
Heinrich Kiliani: On the Composition and Constitution of
Arabinosecarboxylic Acid and of Arabinose
1887
In a report dated 27 November 1886 I showed on one hand that arabaric
acid formed by oxidation of arabinose has the formula C5H10O6, but on the
other hand described several derivatives of arabinose carboxylic acid with
the formula C7H14O8, since at that time I had no basis, in truth, to dispute
the generally accepted formula for arabinose – C6H12O6 – and my analytical
results did not contradict either assumption.
H+
HCN
Na/Hg
H2O
arabinose
mixture
arabinose
carboxylic acid
mannitol
Kiliani-Fischer Synthesis
elongates an aldose
D
Na/Hg
Na/Hg
Emil Fischer
1852-1919
K-F
Phenylhydrazine - Fischer’s First Chemical Love
Osazone
crystalline!
Emil Fischer
1852-1919
The D-Aldoses
D-Glyceraldehyde
Tetroses
Erythrose
Pentoses
Arabinose
Xylose
Hexoses
Glucose
Mannose
Galactose
Gulose
Kiliani
HC=O
Must differ in configuration
How many are there?
Abbreviate
H
HO
H
H
H
OH
H
OH
OH
H
OH
Which is which?
as
Fischer’s Evidence
HNO3
1) D
1) Glucose
Glucaric Diacid 2) Na (Hg) “Gulose”
1a) Glucaric Diacid is chiral (both enantiomers known)
2) Glucose & Mannose give same Osazone;
Arabinose Kiliani Gluconic & Mannonic acids
Fructose Na (Hg) Glucitol & Mannitol
2a) Mannitol & Mannonic Acid are chiral
K-F syn
3) Arabinose
Glucose (& Mannose)
K-F syn
Xylose
Gulose (& Idose?)
3a) Arabinose gives active Arabitol and Arabaric Diacid
Xylose gives inactive Xylitol and Xylaric Diacid
The D-Aldoses
1) Glucose
HNO3
Glucaric Diacid
1) D
2) Na (Hg)
“Gulose”
1a) Glucaric Diacid is chiral (both enantiomers known)
Pentoses
Arabinose
Xylose
Hexoses
Glucose
Mannose
Galactose
Gulose
1)
1a)
•
X
X
X
•
X
The D-Aldoses
2) Glucose & Mannose give same Osazone
Arabinose Kiliani Gluconic & Mannonic acids
Fructose Na (Hg) Glucitol & Mannitol
2a) Mannitol & Mannonic Acid are chiral
Pentoses
Arabinose
Xylose
Hexoses
Glucose
Mannose
Galactose
Gulose
1)
2)
1a)
2a)
•
X
M
X
X
X
M
M
•
X
M
X
The D-Aldoses
K-F syn
3) Arabinose
Glucose (& Mannose)
K-F syn
Xylose
Gulose (& Idose?)
3a) Arabinose gives active Arabitol and Arabaric Diacid
Xylose gives inactive Xylitol and Xylaric Diacid
Pentoses
Arabinose
Xylose
Hexoses
Glucose
Mannose
Galactose
Gulose
1)
2)
3)
1a)
2a)
3a)
•
X
M
X
X
X
M
M
X
•
X
M
X
The D-Aldoses
D-Glyceraldehyde
Tetroses
Erythrose
Erythrose
Threose
Pentoses
Arabinose
Xylose
Ribose
Arabinose
Xylose
Lyxose
Hexoses
Glucose
Mannose
Galactose
Gulose
Allose
Gulose
Glucose
Galactose
Altrose
Mannose
Idose
Talose
All Altruists Gladly Make Gum
In
Gallon Tanks
REVIEW 1:
The Synthesis of Two
Unnatural Products
(in order to settle a question in the
theory of organic chemistry)
Is cyclobutadiene antiaromatic (4n)?
O
O
h
O
(must be disrotatory)
O
h
Make it and see.
Presumptive Evidence of its Existence.
Spectroscopy?
(2 +2 forbidden thermally)
DielsAlder
(2 +2 forbidden thermally, but
it happens anyway)
+ O=C=O
very strained
Make one molecule
per cage
Making & Studying
“antiaromatic”
Cyclobutadiene
mouth
O
CH
CH2
CH2
(for solubility)
Ph
Cram, Tanner, and Thomas (1991)
Preparing
Dihydrocinnamaldehyde
O
O
O
CH
CH
CH2
CH
CH2
CH
O
Ph
Ph
CH
CH
CH3
Ph
1) “Br+” / -H+ (3 moles)
bonds
by SN2
2-Owith
mixture
tetra2)(asO-CH
-
Start with Hemisphere
Ar-O with
2BrCl
substituted
andCH
two
disubstituted analogues)
-
Br
ClCH2
-
Br
Br
Br
How to
form the
C16 ring?
(by chromatography;
5%
from tetramer)
(OH
is activating,
Resorcinol
Hydrocinnamaldehyde
(from benzaldehyde
see above)
o,p-directing)
H+
OH
H+
+
etc. etc.
The electrophilic aromatic
substitution is reversible, and
ultimately the desired “tetramer”
stereoisomer precipitates from the
equilibrating mixture in 69% yield
based on hydrocinnamaldehyde.
Lucky!
1) Br+/-H+ (3 moles)
2) O-CH2--O bonds by SN2
Ar-O with CH2BrCl
3)
Joining Hemispheres
exchange
BuLi (halogen-metal
 more stable “Ar- anion”)
-
-
Li
Br
OH
B(OR)2 Br
Li
OH
B(OR)2
~40%
HO
- -overall)
HO(1%
HO
O
O-B(OR)
2
O-Li
Br
OH
B(OR)
2
4) B(OR)3
(add “Ar- ” to B ; lose RO-)
5) HOO-
(insert O between C and B.
Cf. hydroboration/oxidation;
lose most stable ArO- anion)
Note: the purpose of 1,3,4,5 is to “hide” an OH group
between the OH groups of resorcinol, and then reveal it.
6) O-CH2--O bonds by SN2
Ar-O with CH2BrCl
CHCl3
CHCl3
Stereo Pair
X-Ray View
JACS, 113, 7717 (1991)
(easier to see without a
viewer if you make it small)
CH3CN
CH3CN
CH3CN
CH3CN
CHCl3
CHCl3 &
CH3CN are
held between
adjacent
molecules in
crystal
O
HC-N(CH3)2 held
within molecule.
but lost with
CHCl3
t1/2 = 34 hrs
at 140°C.
Replace DMF
by a-Pyranone
O
O
. . O
. .O
. .
. .
O
O
. .
. .
Most
shift comes from
Antiaromatic
other rings, still ~1.5
upfield
ppm above shift?
benzene
.
Proton NMR
Normal
Benzene as guest
above center of 8 benzene rings
.
.
.. .
.
..
End of Lecture 74
April 27, 20101
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J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0
The Benzoin Condensation (prob. 19.90)
CN “reverses the
polarity” of O=C+
to C- (“umpolung”)
also an
a-activator
(benzylic)
OH
like C=O an
what we have:
Ph C C N a-activator
leaving
N
C
H
O
H C
H Ph
group
C N base
Ph C C N
not basic enoughO
to
C N nucleophile
C N
pull off H+. D pKa > 30
H C Ph
OH
where we’re going:
O
need Ph-C
to attack O=CH-Ph
H+